Question

in a recent poll 45% of survey respondents said that if they would prefer the child to be a boy. suppose you conducted a survey of 190 randomly selected students on your campus and find that 87 of them would prefer a boy

a) use the normal approximation to the binomial to approximate the probability that in a random sample of 190 students at least 87 would prefer a boy assuming the true percentage is 45%

b) does this result contradict the poll? explain

Answer 1

a)

X ~ Bin ( n , p)

Where n = 190 , p = 0.45

Using Normal Approximation to Binomial       
Mean = n * P = ( 190 * 0.45 ) = 85.5      
Variance = n * P * Q = ( 190 * 0.45 * 0.55 ) = 47.025      
Standard deviation = √(variance) = √(47.025) = 6.8575      
      
P(X < x) = P(Z < ( x - mean ) / SD)  

P(X >= 87) = P(X > 86.5)

P ( X > 86.5 ) = P(Z > (86.5 - 85.5 ) / 6.8575 )
= P ( Z > 0.15 )
= 1 - P ( Z < 0.15 )
= 1 - 0.5596
= 0.4404

b)

Since this probability is greater than 0.05, the result is not unusual.

the result does not contradict the poll.